The silicon substrate and adding to - PowerPoint Presentation

Slide1

The silicon substrate and adding to it—Part 1

Explain

how single crystalline Si wafers are made

Describe

the crystalline structure of Si

Find

the

Miller indices

of a planes and directions in crystals and give the most important direction/planes in silicon

Use

wafer flats to identify types of Si wafers

Define

Semiconductor

Doping/dopant

Resistivity

Implantation

Diffusion

p-n junction

Give

a number of uses of p-n

junctions

Calculate

Concentration

distributions for

thermal

diffusion

Concentration

distributions for

ion implantation

,

and

p-n

junction

depths Slide2

Silicon—The big green Lego®

Bulk

micromachining

Surface

micromachining

silicon

substrate

silicon

substrateSlide3

Three forms of material

Crystalline

Polycrystalline

Amorphous

Grains

Silicon wafers

Polysilicon

(in surface

μ

-machining)

Glass and fused quartz, polyimide, photoresistSlide4

Creating silicon wafers

The

Czochralski

method

Creates crystalline (

cristalino

) Si of high purity

A “seed” (

semilla

) of solid Si is placed in molten Si—called the melt—which is then slowly spun and drawn upwards while cooling it. Crucible and the “melt” turned in opposite directions

Wafers cut from the cross section. Slide5

Creating silicon wafers

Photo (

foto

) of a

monocrystalline

silicon ingot

Polycrystalline silicon

(American Ceramics Society)

GrainsSlide6

It’s a crystal

Cubic

Body-centered cubic (BCC)

Face-centered cubic (FCC)

a

Unit cells

a

-

lattice constant

, length of a side of a unit cell

a

aSlide7

It’s a crystal

The diamond (diamante) latticeSlide8

Miller indices

The Miller indices give us a way to identify different directions and planes in a crystalline structure.

Indices:

h

,

k

and

l

[

h k l ]  a specific direction in the crystal

<h k l >



a family of equivalent

directions

(

h k

l

)



a specific

plane

{h k l } 

a family of equivalent planes

How to find Miller indices:

Identify where the plane of interest intersects the three axes forming the unit cell. Express this in terms of an integer multiple of the lattice constant for the appropriate axis.

Next, take the reciprocal of each quantity. This eliminates infinities

.

Finally, multiply the set by the least common denominator. Enclose the set with the appropriate brackets. Negative quantities are usually indicated with an over-score above the number.Slide9

Te toca

a ti

Find the

Miller indices

of the plane shown in the figure.

How to find Miller indices:

Identify

where the plane of interest intersects the three axes forming the unit cell. Express this in terms of an integer multiple of the lattice constant for the appropriate axis

.

Next, take the reciprocal of each quantity. This eliminates infinities

.

Finally, multiply the set by the least common denominator. Enclose the set with the appropriate brackets. Negative quantities are usually indicated with an over-score above the number.

a

b

c

1

2

2

1

1

2

3

4

Respuesta

: (2 4 1)

For

cubic crystals

the Miller indices represent a direction vector perpendicular to a plane with integer components

.

Es

decir

,

[

h k l]

⊥ (h k l)

¡Ojo!

Not

true

for

non-

cubic

materials

!Slide10

Non-cubic material example

Quartz is an example of an important material with a non-cubic crystalline structure.

(http

://

www.jrkermode.co.uk/quippy/adglass.html)Slide11

Miller indices

What are the Miller indices of the shaded planes in the figure below?

(1 0 0)

(1 1 0)

(1 1 1)

Te

toca

a ti

:

Find the angles between

{1 0 0} and {1 1 1} planes, and

{1 1 0} and {1 1 1} planes.Slide12

Wafer types

Si wafers differ based on the orientation of their crystal

planes

in

relation to the surface plane of the

wafer.

Wafers “flats” are used to identify

the crystalline orientation of the surface plane, and

whether the wafer is

n-type

or

p-type

.

(1 0 0) wafer

<1 0 0> directionSlide13

Relative position of crystalline planes in a (100) wafer

Orientations of various crystal directions and planes in a (100) wafer (Adapted from

Peeters

, 1994) Slide14

It’s a semiconductor

Conductors

Insulators

Semiconductors

The “jump” is affected by both temperature and light



sensors and optical switchesSlide15

Conductivity, resistivity, and resistance

Electrical conductivity (σ

)



A measure

of how easily

a material conducts electricity

Material property

Electrical resistivity (ρ) 

Inverse of conductivity; es decir

ρ

= 1/

σ

Material property

By

doping

, the

resistivity of silicon can be varied

over a range of about 1×10

-4

to 1×10

8

Ω

•m!Slide16

Conductivity, resistivity, and resistance

Te

toca

a ti

Find the total resistance (in

Ω

) for the MEMS snake (

serpiente) resistor shown in the figure if it is made of

Aluminum (ρ = 2.52×10-8

Ω·m) andSilicon

100

μ

m

1

μ

m

1

μ

m

Entire resistor is 0.5

μ

m thick

100 bends total

Respuesta

:

Al: 509

Ω

Si: 1.3 G

Ω

!!Slide17

Doping

Phosphorus is a donor

– donates electrons

Boron is an

acceptor

– accepts electrons from Si

 Charge carriers are “holes.”

Phosphorus and boron are both

dopants

.

P creates an

n-type

semiconductor.

B creates a

p-type

semiconductor.Slide18

Doping

Two major methods

Build into wafer itself during silicon growth

Gives a uniform distribution of dopant



Background concentration

Introduce to existing wafer

Implantation

or

diffusion (or both!)

Non-uniform distribution of dopantUsually the opposite type of dopant (Es

decir

,

si

wafer

es

p-type, el

otro

es

n-type y vice versa)

Location where dopant concentration matches background concentration se llama p-n junction

p-n junction

Uses of doping and p-n junctions:

Change electrical properties (make more or less conductive)

Create

piezoresistance

,

piezoelectricity

, etc. to be used for sensing/actuationCreate an etch stopSlide19

Doping

Often implantation and diffusion are done through masks

in

the

wafer surface

in order to create p-n junctions at specific locations.

How do we determine the distribution of diffused and/or implanted dopant?

Mass diffusion:

Mass “flux”

Concentration gradient

Diffusion constant

Compare to

Frequency factor and activation energy for diffusion of dopants in siliconSlide20

Doping by diffusion

Conservation of mass (applied to any point in the wafer)

C

x

time

Need

1 initial condition

2 boundary conditions

At

t

= 0, or

C

(

x

,

t

= 0) = 0

C

(

x

→

∞

,

t

> 0) = 0

C

(

x

= 0,

t

> 0) =

C

s

Solución



erfc

(

) is the complementary error

function:

Appendix CSlide21

Doping by diffusion

Diffusion of boron in silicon at 1050°C for various times

x

Diffusion length

 rough estimate of how far dopant has penetrated waferSlide22

Doping by diffusion

Total amount of dopant diffused into a surface per unit area is called the

ion dose

.

x

time

C

(

x

= 0,

t

> 0) =

C

s

Q = constant

C

s

Gaussian distributionSlide23

Doping by implantation

Distribution is also Gaussian, but it is more complicated.

Doping by ion implantation

C

P

– peak

concentration of

dopant

R

P

–

the

projected

range

(the

depth of peak concentration of dopant in wafer

)

Δ

R

P

– standard deviation of the distributionRange affected by the mass of the dopant, its acceleration energy, and the stopping power of the substrate material.

Peak concentration

 Slide24

Doping by implantation

Doping is often (de hecho, usually) a two-step process:

1

st

implantation –

pre-deposition

2

nd

thermal diffusion – drive-in If projected range of pre-deposition is small, can approximate distribution with

Typical concentration profiles for ion implantation of various dopant species

Replace with

Q

iSlide25

Junction depth

x

C

implanted dopant

background concentration

p-n junctionSlide26

Te toca

a ti

A n-type

Si-wafer with background doping

concentration

of

2.00×10

15

cm-3 is doped by ion implantation with a dose of

boron atoms of 1015 cm

-2, located on the surface of the wafer. Next thermal diffusion is used for the drive-in of boron atoms into the wafer a 900°C for

4

hours.

a. What is the diffusion constant of

boron in silicon at this

temperature?

b. What is the junction depth after

drive-in?

Hints:

Assume that the distribution of ions due to implantation is very close to the wafer surface

Useful information:

k

b = 1.381×10-23

J/KeV = 1.602×10-19 JSlide27

Te toca

a ti

A n-type

Si-wafer with background doping

concentration

of

2.00×10

15

cm-3 is doped by ion implantation with a dose of

boron atoms of 1015 cm

-2, located on the surface of the wafer. Next thermal diffusion is used for the drive-in of boron atoms into the wafer a 900°C for

4

hours.

a. What is the diffusion constant of

boron in silicon at this

temperature?

b. What is the junction depth after

drive-in?

Replace with

Q

i

Set =

C

bg

a.

b.

=

1.248×10

-18

m

2

/s

= 0.83×10-6 m